Schaum's Outline of Advanced Calculus (3rd Edition) (Schaum's Outlines Series)

Schaum's Outline of Advanced Calculus (3rd Edition) (Schaum's Outlines Series)

Murray R. Spiegel, Robert Wrede

Language: English

Pages: 456

ISBN: B01K0UJ1NQ

Format: PDF / Kindle (mobi) / ePub


Tough Test Questions? Missed Lectures? Not Enough Time?

Fortunately for you, there's Schaum's.

More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Schaum's is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.

This Schaum's Outline gives you

1,370 fully solved problems
Complete review of all course fundamentals
Clear, concise explanations of all Advanced Calculus concepts
Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time--and get your best test scores!

Topics include: Numbers; Sequences; Functions, Limits, and Continuity; Derivatives; Integrals; Partial Derivatives; Vectors; Applications of Partial Derivatives; Multiple Integrals; Line Integrals, Surface Integrals, and Integral Theorems; Infinite Series; Improper Integrals; Fourier Series; Fourier Integrals; Gamma and Beta Functions; and Functions of a Complex Variable

Schaum's Outlines--Problem Solved.

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is any positive number, we define lim inf fun g ¼ À1. If lim un ¼ 1, we define lim sup fun g ¼ lim inf fun g ¼ 1. n!1 If lim un ¼ À1, we define lim sup fun g ¼ lim inf fun g ¼ À1. n!1 Although every bounded sequence is not necessarily convergent, it always has a finite lim sup and lim inf. A sequence fun g converges if and only if lim sup un ¼ lim inf un is finite. NESTED INTERVALS Consider a set of intervals ½an ; bn Š, n ¼ 1; 2; 3; . . . ; where each interval is contained in the preceding one and

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi dx cos y 2 1 À x2 1 À sin y We have supposed here that the principal value À=2 @ sinÀ1 x @ =2, is chosen so that cos y is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi positive, thus accounting for our writing cos y ¼ 1 À sin2 y rather than cos y ¼ Æ 1 À sin2 y. 4.15. Derive the formula d loga e du ðloga uÞ ¼ ða > 0; a 6¼ 1Þ, where u is a differentiable function of x. dx u dx Consider y ¼ f ðuÞ ¼ loga u. By definition, dy f ðu þ ÁuÞ À f ðuÞ loga ðu þ ÁuÞ À

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A21 þ A22 þ A23 . See Fig. CHAP. 7] 165 VECTORS C b A 1 b 2 D 1 a 2 d E a c B Fig. 7-19 Fig. 7-20 By the Pythagorean theorem, ðOPÞ2 ¼ ðOQÞ2 þ ðQPÞ2 Similarly, ðOQÞ2 ¼ ðORÞ2 þ ðRQÞ2 . qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Then ðOPÞ2 ¼ ðORÞ2 þ ðRQÞ2 þ ðQPÞ2 or A2 ¼ A21 þ A22 þ A23 , i.e., A ¼ A21 þ A22 þ A23 . where OP denotes the magnitude of vector OP, etc. 7.9. Determine the vector having initial point Pðx1 ; y1 ; z1 Þ and terminal point Qðx2 ; y2 ; z2 Þ

curvilinear coordinates, rÈ ¼ e1 @È e2 @È e3 @È þ þ h1 @u1 h2 @u2 h3 @u3 [Hint: Let rÈ ¼ a1 e1 þ a2 e2 þ a3 e3 and use the fact that dÈ ¼ rÈ Á dr must be the same in both rectangular and the curvilinear coordinates.] 7.89. Give a vector interpretation to the theorem in Problem 6.35 of Chapter 6. MISCELLANEOUS PROBLEMS 7.90. If A is a differentiable function of u and jAðuÞj ¼ 1, prove that dA=du is perpendicular to A. 7.91. Prove formulas 6, 7, and 8 on Page 159. 7.92. If  and  are polar

vector r À r0 ¼ ðx À 1Þi þ ð y À 2Þj þ ðz þ 1Þk lies in the tangent plane and is thus perpendicular to N0 . Then the required equation is ðr À r0 Þ Á N0 ¼ 0 i:e:; fðx À 1Þi þ ð y À 2Þj þ ðz þ 1Þkg Á fÀ6i þ 11j þ 14kg ¼ 0 À6ðx À 1Þ þ 11ð y À 2Þ þ 14ðz þ 1Þ ¼ 0 6x À 11y À 14z þ 2 ¼ 0 or (b) Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal N0 . The vector from O to the point ð1; 2; À1Þ on the normal is r0 ¼ i þ 2j À k. The vector r À r0 ¼ ðx À 1Þi þ ð y À 2Þj þ ðz

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